16 votes
Accepted

The Lévy-Khintchine formula and integrability conditions of a random measure

The condition $$\int_{\mathbb{R} \backslash \{0\}} \min\{1,z^2\} \nu(dz)<\infty$$ is equivalent to $$\int_{|z| \leq 1} z^2 \, \nu(dz) < \infty \quad \text{and} \quad \int_{|z| \geq 1} \nu(dz) ...
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11 votes
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Proving properties for the Poisson-process.

Let $(X_t)_{t \geq 0}$ be a Poisson process with intensity $\lambda$. Step 1: $(X_t)_{t \geq 0}$ has almost surely increasing sample paths. Proof: Fix $s \leq t$. Since $(X_t)_{t \geq 0}$ has ...
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  • 113k
10 votes
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Jumps of Lévy process

Yes, there is a very strong relation between the (distribution of the) jumps of a Lévy process and its Lévy measure. In fact, the Lévy measure describes the jump behaviour of the corresponding Lévy ...
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  • 113k
7 votes

Good book that contains stochastic integration, martingales and Lévy-processes?

Since Lévy processes are used a lot in finance, there are several books on this topic (that is, Lévy processes and their applications in finance). For example "Stochastic Calculus for Finance II" by S....
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7 votes
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Etemadi's inequality

The inequality which both ziT and I used is a direct consequence of the following inequality. Lemma 1: Let $X_1,\ldots,X_n$ be independent random variables and $S_k := \sum_{j=1}^k X_j$, $k=1,\...
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  • 113k
6 votes

Application of Lévy–Khinchine formula

Depending on what you know about the Lévy triplet it is much easier to calculate the characteristic functions using the very definition of Wiener Process and Poisson process, respectively. For example,...
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5 votes
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Stochastic Continuity of a Lévy Process

By the homogeneity of the increments, we have $$\mathbb{P}(|X(s)-X(t)|>\varepsilon) = \mathbb{P}(|X(t-s)|>\varepsilon).$$ This means that $$\lim_{s \to t} \mathbb{P}(|X(s)-X(t)|>\...
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5 votes
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Largest jumps of a spectrally positive $\alpha$-stable process

Denote by $N$ the jump measure of the Lévy process, i.e. $$N_t(B) := N([0,t] \times B) := \sharp \{s \in [0,t]; \Delta X_s := X_s-X_{s-} \in B\},$$ and by $\nu$ its Lévy measure. It is widely known ...
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5 votes
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Central Limit Theorem for Lévy Process

Without any additional assumptions on the Lévy process $(X_t)_{t \geq 0}$, a central limit theorem does not hold true. Let $(X_t)_{t \geq 0}$ be a (one-dimensional) Lévy process with Lévy triplet $(...
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  • 113k
5 votes
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Gaussian process - Hölder continous paths

The Kolmogorov-Chentsov theorem states that if for any $T\gt 0$, there exists $\alpha,\beta$ and $C\gt 0$ such that for any $s,t\in [0,T ]$, $$ \tag{1} \mathbb E \left |X_t-X_s\right|^ \alpha \...
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4 votes
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Finiteness of the number of big jumps of a Lévy process on a finite interval

Let $(Y_t)_{t \in [0,1]}$ be a Lévy process and denote by $\Delta Y_t := Y_t-Y_{t-}$ the jump height at time $t$. The aim is to show that $(Y_t)_{t \in [0,1]}$ has only finitely many jumps with jump ...
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4 votes
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Lévy-Khintchine formula and Taylor expansion

The condition ensures that the integral $$\int_{\mathbb{R} \backslash \{0\}} (e^{iux}-1-iux 1_{\{|x|<1\}}) \, \nu(dx)$$ is well-defined. Since $|e^{iux}| \leq 1$, we have $$|e^{iux}-1| \leq 2$$...
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  • 113k
4 votes

Left-continuity of a Lévy filtration

Is essentially $\mathcal{F}_t = \mathcal{F}_{t-}+$ null sets? Yes. Roughly speaking, $\mathcal{F}_t$ contains all the information in $\mathcal{F}_{t-}$ plus the information for which $\omega$ a jump ...
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  • 113k
4 votes

Good book that contains stochastic integration, martingales and Lévy-processes?

It depends a little bit on your interests, but as you might know, stochastic processes and Itô-calculus is excessively used in quantitative finance. I can recommend some books which really explain the ...
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  • 1,350
4 votes
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Integration with respect to a Poisson random measure

(1) $\iff$ (2) Since $f \geq 0$ is measurable, there exists a sequence $(f_n)_{n \in \mathbb{N}}$ of simple functions (aka elemtary functions/step functions) such that $0 \leq f_n \uparrow f$. Each $...
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4 votes
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Why is $T_n-T_{n-1}$ independent of $\mathcal{F}_{T_{n-1}}$ where $T_{n}=\inf \{t>T_{n-1}: |X_t-X_{T_{n-1}}| \geq C\}$?

Let $T$ be a stopping time and $Y_t := X_{T+t}-X_T$ the restarted Lévy process. Theorem 32 states that the $\sigma$-algebra generated by $(Y_t)_{t \geq 0}$ is independent of $\mathcal{F}_T$, i.e. $$\...
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  • 113k
4 votes
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What is the true definition of a Lévy process?

Typically one distinguishes between "Lévy processes" and "Lévy processes in law". Definition: Let $(X_t)_{t \geq 0}$ be a stochastic process such that $X_0 = 0$ almost surely and $(X_t)_{t \geq 0}$...
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  • 113k
4 votes
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Why are processes with stationary independent increments nonstationary?

If $(N_t)_{t \geq 0}$ is a Lévy process (i.e. a stochastic process with stationary and independent increments) and $h \neq 0$, then the random variables $N_t$ and $N_{t+h}$ cannot have the same ...
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  • 113k
3 votes
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Change from stochastic exponential to exponential of Lévy process - Applebaum

$\nu \circ f^{-1}$ denotes the push-forward measure (or image measure) of $f$ with respect to $\nu$, i.e. $$\nu_1(B) = (\nu \circ f^{-1})(B) = \int 1_B(f(x)) \, \nu(dx)$$ for any Borel set $B$. This ...
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  • 113k
3 votes
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Quadratic Variation of a square-integrable Lévy process

Hints: Since $(X_t)_{t \geq 0}$ is a martingale and $X_0=0$, we have $\mathbb{E}(X_t)=0$ for all $t \geq 0$. If a random variable $Y$ has finite second moment, then $$\mathbb{E}(Y) = \frac{1}{\imath} ...
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  • 113k
3 votes
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Intuition behind a set of a assumptions about Lévy processes.

This assumptions means in particular that the jump measure $\nu$ of the Lévy process $(X_t)_{t \geq 0}$ has a density $w$ with respect to the Lebesgue measure. Moreover, the form of the integral shows ...
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3 votes
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Blumenthal-Getoor index for IG and $\Gamma$

Example 1 (Gamma process): As @Goulifet pointed out, the Lévy measure of a Gamma process is given by $$\nu(dx) =c x^{-1} e^{-\alpha x} \, dx =: p(x) \, dx.$$ Here $\alpha$ and $c$ are fixed constants. ...
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  • 113k
3 votes
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$E[1_{\lbrace P_T-P_{\tau_n}=0\rbrace}\int_{\tau_n}^T h(s)dN_s]=0?$

The answer to the first question is yes : $X_t=\int_{t}^T h(s)dN_s$ this process in not adapted to the natural filtration of $N$ (as it looks into the future) so we can not apply optiaml sampling ...
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  • 5,417
3 votes

Random process with stationary independent increments determined by first order distribution?

Note that the equality $$X_t = 2 X_{t/2}$$ does not hold. It follows from stationarity of the increments that $$X_{t}-X_{t/2} \stackrel{d}{\sim} X_{t/2},$$ i.e. they are equal in distribution and ...
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  • 113k
3 votes
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Subordination of a Levy process when the "subordinator" is not nondecreasing

Usually, Lévy processes $(X_t)_{t \geq 0}$ are considered, i.e. the time $t$ is non-negative. This means that we need the subordinator $(Z_t)_{t \geq 0}$ to satisfy $Z_t \geq 0$. Since Lévy processes ...
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  • 113k
3 votes
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Lévy Process existence of the expectation of the supremum of the past process.

Yeah, applying Etemadi's inequality is a good idea. The following identity, which holds for any non-negative random variable $X$, will also be useful: $$\mathbb{E}(X) = \int_0^{\infty} \mathbb{P}(X &...
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  • 113k
3 votes

Stochastic Integral with respect to Compensated Poisson Process

This might be helpful: Theorem 29 on page 173 of Protter's "Stochastic integration and differential equations": Let $M$ be a local martingale and let $H\in \mathcal P$ be locally bounded. then the ...
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  • 1,088
3 votes
Accepted

Lévy-process property

Yes, this results holds for any Lévy process which is integrable. It follows from the independence and stationarity of the increments that $$\mathbb{E}X_{k/n} = \mathbb{E} \left( \sum_{j=1}^k (X_{j/...
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  • 113k
3 votes

Why is a continuous Lévy process twice integrable?

Let me try to indicate an answer to what I think is your real question ("Is there a direct way to show that a continuous Lévy process is square integrable?") A detailed discussion of this and much ...
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  • 22.8k
3 votes
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Ornstein-Uhlenbeck Process simulation bug

If the time vector is t = 0:dt:500; there is an overflow in exp(th*t(i+1)) in the for loop, as t gets large. Perhaps something like this: ...
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  • 1,312

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